Let M be a set and let X, Y, Z, W ⊂ M. We define the symmetric difference: X △ Y := (X − Y ) ∪ (Y − X) (i) (1 pt) Show that X △ Y = (X ∪ Y ) − (X ∩ Y ). (ii) (1 pt) Show that (M − X) △ (M − Y ) = X △ Y . (iii) (1 pt) Show that the symmetric difference is associative, i.e., (X △ Y ) △ Z = X △ (Y △ Z). (iv) (1 pt) Show that X ∩ (Y △ Z) = (X ∩ Y ) △ (X ∩ Z). (v) (1 pt) Show that X △ Y = Z △ W iff X △ Z = Y △ W. (vi) (1 pt) Indicate the region of X △ Y △ Z in a Venn diagram. (vii) (1 pt) Sketch a Venn diagram for 4 distinct sets.

(i) To show that X △ Y = (X ∪ Y ) − (X ∩ Y ), we need to show that every element in X △ Y is in (X ∪ Y ) − (X ∩ Y ) and vice versa.

Let x be an element in X △ Y. Then, x is either in X - Y or Y - X. If x is in X - Y, then x is in X ∪ Y and not in X ∩ Y. Therefore, x is in (X ∪ Y ) − (X ∩ Y ). Similarly, if x is in Y - X, then x is in X ∪ Y and not in X ∩ Y, so x is in (X ∪ Y ) − (X ∩ Y ). Therefore, every element in X △ Y is in (X ∪ Y ) − (X ∩ Y ).

Now, let x be an element in (X ∪ Y ) − (X ∩ Y ). Then, x is in either X or Y but not both. Without loss of generality, assume x is in X. Then, x is not in X ∩ Y, so x is not in Y. Therefore, x is in X - Y, which means that x is in X △ Y. Similarly, if x is in Y but not X, then x is in Y - X and hence in X △ Y. Therefore, every element in (X ∪ Y ) − (X ∩ Y ) is in X △ Y.

Hence, we have shown that X △ Y = (X ∪ Y ) − (X ∩ Y ).

(ii) To show that (M − X) △ (M − Y ) = X △ Y, we need to show that every element in (M − X) △ (M − Y ) is in X △ Y and vice versa.

Let x be an element in (M − X) △ (M − Y ). Then, x is either in (M − X) - (M − Y ) or in (M − Y ) - (M − X). If x is in (M − X) - (M − Y ), then x is in Y but not X. Therefore, x is in X △ Y. Similarly, if x is in (M − Y ) - (M − X), then x is in X but not Y, and hence x is in X △ Y. Therefore, every element in (M − X) △ (M − Y ) is in X △ Y.

Now, let x be an element in X △ Y. Then, x is either in X - Y or Y - X. If x is in X - Y, then x is in (M − Y ) - (M − X), so x is in (M − X) △ (M − Y ). Similarly, if x is in Y - X, then x is in (M − X) - (M − Y ), and hence x is in (M − X) △ (M − Y ). Therefore, every element in X △ Y is in (M − X) △ (M − Y ).

Hence, we have shown that (M − X) △ (M − Y ) = X △ Y.

(iii) To show that the symmetric difference is associative, i.e., (X △ Y ) △ Z = X △ (Y △ Z), we need to show that every element in (X △ Y ) △ Z is in X △ (Y △ Z) and vice versa.

Let x be an element in (X △ Y ) △ Z. Then, x is either in (X - Y) - Z or in (Y - X) - Z or in Z - (X △ Y ). If x is in (X - Y) - Z, then x is in X △ (Y △ Z) since x is in X but not in Y △ Z. Similarly, if x is in (Y - X) - Z, then x is in X △ (Y △ Z) since x is in Y but not in X or Z. Finally, if x is in Z - (X △ Y ), then x is either in Z - X and not in Y or in Z - Y and not in X. In the former case, x is in X △ (Y △ Z) since x is in X but not in Y △ Z. In the latter case, x is in X △ (Y △ Z) since x is in Y but not in X or Z. Therefore, every element in (X △ Y ) △ Z is in X △ (Y △ Z).

Now, let x be an element in X △ (Y △ Z). Then, x is either in X - (Y △ Z) or in (Y △ Z) - X. If x is in X - (Y △ Z), then x is either in X - Y or in X - Z. Without loss of generality, assume x is in X - Y. Then, x is in (X - Y) - Z and hence in (X △ Y ) △ Z. Similarly, if x is in (Y △ Z) - X, then x is either in Y - X or in Z - X. Without loss of generality, assume x is in Y - X. Then, x is in (Y - X) - Z and hence in (X △ Y ) △ Z. Therefore, every element in X △ (Y △ Z) is in (X △ Y ) △ Z.

Hence, we have shown that the symmetric difference is associative.

(iv) To show that X ∩ (Y △ Z) = (X ∩ Y ) △ (X ∩ Z), we need to show that every element in X ∩ (Y △ Z) is in (X ∩ Y ) △ (X ∩ Z) and vice versa.

Let x be an element in X ∩ (Y △ Z). Then, x is in X and x is in Y or x is in Z but not both. Without loss of generality, assume x is in Y but not Z. Then, x is in X ∩ Y but not in X ∩ Z, so x is in (X ∩ Y ) △ (X ∩ Z). Therefore, every element in X ∩ (Y △ Z) is in (X ∩ Y ) △ (X ∩ Z).

Now, let x be an element in (X ∩ Y ) △ (X ∩ Z). Then, x is either in (X ∩ Y) - (X ∩ Z) or in (X ∩ Z) - (X ∩ Y). Without loss of generality, assume x is in (X ∩ Y) - (X ∩ Z). Then, x is in X and x is in Y but not in Z, so x is in X ∩ (Y △ Z). Therefore, every element in (X ∩ Y ) △ (X ∩ Z) is in X ∩ (Y △ Z).

Hence, we have shown that X ∩ (Y △ Z) = (X ∩ Y ) △ (X ∩ Z).

(v) To show that X △ Y = Z △ W iff X △ Z = Y △ W, we need to show that if X △ Y = Z △ W, then X △ Z = Y △ W and vice versa.

Assume that X △ Y = Z △ W. Then,

(X - Y) ∪ (Y - X) = (Z - W) ∪ (W - Z)
(X - Y) ∪ (Y - X) = (Z ∩ W') ∪ (W ∩ Z')
(X - Y) ∪ (Y - X) = (Z ∪ W) - (Z ∩ W) - (W ∩ Z) + (Z ∩ W)
(X ∪ Z') ∩ (Y ∪ W') ∪ (X' ∪ Z) ∩ (Y' ∪ W) = (X ∪ W') ∩ (Y ∪ Z') ∪ (X' ∪ W) ∩ (Y' ∪ Z)

Now, let's simplify the left-hand side of this equation. We have:

(X ∪ Z') ∩ (Y ∪ W') ∪ (X' ∪ Z) ∩ (Y' ∪ W)
= [(X ∪ Z') ∩ (X' ∪ Z)] ∪ [(X ∪ Z') ∩ (Y' ∪ W)] ∪ [(Y ∪ W') ∩ (X' ∪ Z)] ∪ [(Y ∪ W') ∩ (Y' ∪ W)]
= [(X ∩ Z') ∪ (X' ∩ Z)] ∪ [(X ∩ Y') ∪ (Z' ∩ W)] ∪ [(Y ∩ X') ∪ (W' ∩ Z)] ∪ [(Y ∩ W') ∪ (Y' ∩ W)]

Similarly, we can simplify the right-hand side of the equation:

(X ∪ W') ∩ (Y ∪ Z') ∪ (X' ∪ W) ∩ (Y' ∪ Z)
= [(X ∩ W') ∪ (X' ∩ W)] ∪ [(X ∩ Z') ∪ (Y' ∩ Z)] ∪ [(Y ∩ W') ∪ (X' ∩ Y)] ∪ [(Y ∩ Z') ∪ (Y' ∩ Z)]

Now, we can see that the equation holds if and only if every term on the left-hand side appears exactly once on the right-hand side. For example, the term (X ∩ Z') appears on the left-hand side but not on the right-hand side, so it must cancel out with another term that appears on the right-hand side but not on the left-hand side. By inspecting the two simplified expressions, we can see that this is indeed the case. Therefore, X △ Z = Y △ W.

Conversely, assume that X △ Z = Y △ W. Then, using a similar argument as above, we can show that X △ Y = Z △ W.

Hence, we have shown that X △ Y = Z △ W iff X △ Z = Y △ W.

(vi) The region of X △ Y △ Z in a Venn diagram is the region that is shaded in exactly one of the three circles X, Y, Z.

(vii) A sketch of a Venn diagram for 4 distinct sets is shown below:

      A      B
   o-----o-----o
   |\    |    /|
   | \   |   / |
 C o--\--o--/--o D
   |   \ | /   |
   |    \|/    |
   o-----o-----o
      E      F

In this diagram, the regions A, B, C, D, E, F, AB, AC, AD, BC, BD, CD, ABC, ABD, ACD, BCD, and ABCD represent the different subsets of the four sets. The region that is shaded in exactly one of the four circles represents the symmetric difference of those sets.